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# Find the sum of all the terms in the n-th row of the given series

• Difficulty Level : Easy
• Last Updated : 18 Mar, 2021

Find the sum of all the terms in the nth row of the series given below.

```               1  2
3  4  5  6
7  8  9 10 11 12
13 14 15 16 17 18 19 20
..........................
............................
(so on)```

Examples:

```Input : n = 2
Output : 18
terms in 2nd row and their sum
sum = (3 + 4 + 5 + 6) = 18

Input : n = 4
Output : 132```

Naive Approach: Using two loops. Outer loop executes for i = 1 to n times. Inner loop executes for j = 1 to 2 * i times. Counter variable k to keep track of the current term in the series. When i = n, the values of k are accumulated to the sum.
Time Complexity: O(k), where k is the total number of terms from the beginning till the end of the nth row.
Efficient Approach: The sum of all the terms in the nth row can be obtained by the formula:

` Sum(n) = n * (2 * n2 + 1)`

The proof for the formula is given below:
Prerequisite:

1. Sum of n terms of an Arithmetic Progression series with a as the first term and d as the common difference is given as:

`   Sum = (n * [2*a + (n-1)*d]) / 2`
1.
2. Sum of 1st n natural numbers is given as:

`   Sum = (n * (n + 1)) / 2`
1.

Proof:

```Let the number of terms from the beginning
till the end of the nth row be p.
Here p = 2 + 4 + 6 + .....n terms
For the given AP series, a = 2, d = 2.
Using the above formula for the sum of
n terms of the AP series, we get,

p = n * (n + 1)

Similarly, let the number of terms from the
beginning till the end of the (n-1)th row be q.
Here q = 2 + 4 + 6 + .....n-1 terms
For the given AP series, a = 2, d = 2.
Using the above formula for the sum of
n-1 terms of the AP series, we get,

q = n * (n - 1)

Now,
Sum of all the terms in the nth row
= sum of 1st p natural numbers -
sum of 1st q natural numbers

= (p * (p + 1)) / 2 - (q * (q + 1)) / 2

Substituting the values of p and q and then solving
the equation, we will get,

Sum of all the terms in the nth row = n * (2 * n2 + 1)```

## C++

 `// C++ implementation to find the sum of all the``// terms in the nth row of the given series``#include ` `using` `namespace` `std;` `// function to find the required sum``int` `sumOfTermsInNthRow(``int` `n)``{``    ``// sum = n * (2 * n^2 + 1)``    ``int` `sum = n * (2 * ``pow``(n, 2) + 1);``    ``return` `sum;``}` `// Driver program to test above``int` `main()``{``    ``int` `n = 4;``    ``cout << ``"Sum of all the terms in nth row = "``         ``<< sumOfTermsInNthRow(n);``    ``return` `0;``}`

## Java

 `// Java implementation to find the sum of all the``// terms in the nth row of the given series` `import` `static` `java.lang.Math.pow;` `class` `Test {``    ``// method to find the required sum``    ``static` `int` `sumOfTermsInNthRow(``int` `n)``    ``{``        ``// sum = n * (2 * n^2 + 1)``        ``int` `sum = (``int``)(n * (``2` `* pow(n, ``2``) + ``1``));``        ``return` `sum;``    ``}` `    ``// Driver method``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `n = ``4``;``        ``System.out.println(``"Sum of all the terms in nth row = "``                           ``+ sumOfTermsInNthRow(n));``    ``}``}`

## Python3

 `# Python 3 implementation to find``# the sum of all the terms in the``# nth row of the given series``from` `math ``import` `pow` `# function to find the required sum``def` `sumOfTermsInNthRow(n):``    ` `    ``# sum = n * (2 * n^2 + 1)``    ``sum` `=` `n ``*` `(``2` `*` `pow``(n, ``2``) ``+` `1``)``    ``return` `sum` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `4``    ``print``(``"Sum of all the terms in nth row ="``,``                   ``int``(sumOfTermsInNthRow(n)))` `# This code is contributed``# by Surendra_Gangwar`

## C#

 `// C# implementation to find the sum of all the``// terms in the nth row of the given series``using` `System;` `class` `Test {``    ``// method to find the required sum``    ``static` `int` `sumOfTermsInNthRow(``int` `n)``    ``{``        ``// sum = n * (2 * n^2 + 1)``        ``int` `sum = (``int``)(n * (2 * Math.Pow(n, 2) + 1));``        ``return` `sum;``    ``}` `    ``// Driver method``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 4;``        ``Console.Write(``"Sum of all the terms in nth row = "``                      ``+ sumOfTermsInNthRow(n));``    ``}``}` `// This code is contributed by vt_m.`

## PHP

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## Javascript

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Output:

`Sum of all the terms in nth row = 132`

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